In the equation below
1. If a sample containing 12.5 moles of NH3 is reacted with excess CuO, how many moles of each product can be made?
N2=
Cu=
H20=
thanks!
Answers
Answer:
6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.
Explanation:
We are given the chemical equation:
[tex]\displaystyle 2\text{NH$_3$}_\text{(g)} + 3\text{CuO}_\text{(s)} \longrightarrow \text{N$_2$}_\text{(g)} + 3\text{Cu}_\text{(s)}+3\text{H$_2$O}_\text{(g)}[/tex]
And we want to determine the amount of products produced when 12.5 moles of NH₃ is reacted with excess CuO.
Compute using stoichiometry. From the equation, we can see the following stoichiometric ratios:
The ratio between NH₃ and N₂ is 2:1. (i.e. One mole of N₂ is produced from every two moles of NH₃.)The ratio between NH₃ and Cu is 2:3. The ratio between NH₃ and H₂O is 2:3. (i.e. Three moles of H₂O or Cu is produced frome every two moles of NH₃.)Dimensional Analysis:
The amount of N₂ produced:[tex]\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{1\text{ mol N$_2$}}{2\text{ mol NH$_3$}} = 6.25\text{ mol N$_2$}[/tex]
The amount of Cu produced:[tex]\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{3\text{ mol Cu}}{2\text{ mol NH$_3$}} = 18.8\text{ mol Cu}[/tex]
And the amount of H₂O produced:[tex]\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{3\text{ mol H$_2$O}}{2\text{ mol NH$_3$}} = 18.8\text{ mol H$_2$O}[/tex]
In conclusion, 6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.
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Answers
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Answers
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Urgent!!!
Type the correct answer in the box. Express your answer to three significant figures.
The half-life of carbon-14 is 5,730 years. Dating organic material by looking for C-14 can’t be accurately done after 50,000 years.
Suppose a fossilized tree branch originally contained 4.30 grams of C-14. How much C-14 would be left after 50,000 years?
Use the formula N = N0(1/2)^t/T .
A tree branch that originally had 4.3 grams of carbon-14 will have _____ grams after 50,000 years.
Answers
Answer:
0.0102 g
Explanation:
A common formula for determining the amount of sample remaining in terms of its half-life is
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n = 8.726
===============
N₀ = 4.30 g Calculate N
N = 4.30 × 2.36 × 10⁻³
N₀ = 0.0102 g
Explanation:
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